原式 ax + by = c    =>  ax1 + by1 = gcd(a,b);

a,b,c为任意整数，d = gcd(a,b),则  ax1 + by1 = d 的一组解是（x1,y1）,c是gcd(a,b)的倍数时，其中的一组解为（x1*c/d,y1*c/d）;c不是gcd(a,b)的倍数时，无解

青蛙的约会，就是一道例题

按照题意很容易列举出等式：(x+ms) - (y+ns) = k*l;  (k=1.....n)   变形到  扩展欧几里得公式  即可；

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
using namespace std;
#define MIN INT_MIN
#define MAX INT_MAX
#define N 204
#define LL long long
int gcd(int n,int m)
{
    int r;
    while(m!=0)
    {
        r = n%m;
        n = m;
        m = r;
    }
    return n;
}
void exgcd(LL a,LL b,LL &x1,LL &y1)
{
     if(b==0)
       {
              x1=1;
              y1=0;
              return ;
       }
       exgcd(b,a%b,x1,y1);
       LL t;
       t=x1;
       x1=y1;
       y1=t-a/b*y1;
}
int main()
{
    LL n,m,x,y,s,l,x1,y1;
    scanf("%lld%lld%lld%lld%lld",&x,&y,&m,&n,&l);
    //推导过程如下
    //(x+ms) - (y+ns) = k*l;(k=1.....n)
    //x-y + ms - ns = k*l;
    //(n-m)s + k*l = x-y;
    // a*s + k*b = c; ->gcd(a,b);
    // s为所求最短步数
    LL st = gcd(n-m,l);
    if((x-y)%st!=0) puts("Impossible");
    else
    {
        LL a = n-m;
        LL b = l;
        LL c = x-y;
        a /= st; b /= st; c /= st;
        exgcd(a,b,x1,y1);
        x1 = x1 * c;
       // printf("%lld  %lld\n",x1,y1);
        //y1 = y1 * c;
        //其中一组解(x1*c/st,y1*c/st)->前提c是gcd(a,b)的倍数
       LL int tt = x1 % l;
       //LL int tt = y1*c/st;
        if(tt < 0)
        {
            tt += l;
        }
        printf("%d\n",tt);
    }
    return 0;
}